Sửa đề: $AC=BM$
Có $CK\bot AB, BH\bot AC$
$\Rightarrow \widehat{AKC}=\widehat{AHB}=90^o$
$\Delta AKC$ có $\widehat{ACK}=180^o-\widehat{AKC}-\widehat{BAC}=90^o-\widehat{BAC}$
$\Delta AHB$ có $\widehat{ABH}=180^o-\widehat{AHB}-\widehat{BAC}=90^o-\widehat{BAC}$
Vậy $\widehat{ACK}=\widehat{ABH}$
$\Leftrightarrow 180^o-\widehat{ACK}=180^o-\widehat{ABH}$
$\Leftrightarrow \widehat{ACH}=\widehat{ABM}$
Mà $AB=CN, AC=CM$
$\Rightarrow \Delta ACN=\Delta ABM$ (c.g.c)
$\Rightarrow AM=AN$
